Solve for $r$, $ \dfrac{r + 7}{2r + 1} = -\dfrac{6}{10r + 5} + \dfrac{8}{6r + 3} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2r + 1$ $10r + 5$ and $6r + 3$ The common denominator is $30r + 15$ To get $30r + 15$ in the denominator of the first term, multiply it by $\frac{15}{15}$ $ \dfrac{r + 7}{2r + 1} \times \dfrac{15}{15} = \dfrac{15r + 105}{30r + 15} $ To get $30r + 15$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{6}{10r + 5} \times \dfrac{3}{3} = -\dfrac{18}{30r + 15} $ To get $30r + 15$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{8}{6r + 3} \times \dfrac{5}{5} = \dfrac{40}{30r + 15} $ This give us: $ \dfrac{15r + 105}{30r + 15} = -\dfrac{18}{30r + 15} + \dfrac{40}{30r + 15} $ If we multiply both sides of the equation by $30r + 15$ , we get: $ 15r + 105 = -18 + 40$ $ 15r + 105 = 22$ $ 15r = -83 $ $ r = -\dfrac{83}{15}$